Vibrating alert

We assume, that at time $t = 0$ the unbalance of a vibration motor is aligned along the x-axis. On a mass element $\Delta m$ of the unbalance at the position $\vec r$ a centrifugal force acts due to the rotation around the origin $\Delta \vec F_\text{cf} = \Delta m \omega^2 \vec r = \Delta m \omega^2 \left( \begin{array}{c} r \cos \varphi \\ r \sin \varphi \end{array} \right)$ The total force on the total imbalance results from integration over the semicircular surface: $\begin{aligned} \vec F_\text{cf} &= \int d \vec F_\text{cf} = \int \omega^2 \vec r \,dm = \frac{m}{\frac{1}{2} \pi a^2} \int \omega^2 \vec r \,dA \\ &= \frac{m}{\frac{1}{2} \pi a^2} \int_{0}^{a} \left[\int_{-\pi/2}^{\pi/2} \omega^2 r \left( \begin{array}{c} \cos \varphi \\ \sin \varphi \end{array} \right) \cdot r d\varphi \right] dr \\ &= \frac{m \omega^2}{\frac{1}{2} \pi a^2} \int_{-\pi/2}^{\pi/2} \left( \begin{array}{c} \cos \varphi \\ \sin \varphi \end{array} \right) d\varphi \cdot \int_{0}^{a} r^2 dr \\ &= \frac{m \omega^2}{\frac{1}{2} \pi a^2} \left( \begin{array}{c} 2 \\ 0 \end{array} \right) \cdot \frac{1}{3} a^3 = \frac{4 m \omega^2 a}{3 \pi } \vec e_x \end{aligned}$ As a function of time, there is a centrifugal force $\vec F_\text{cf}(t) = \frac{4 m \omega^2 a}{3 \pi } \left( \begin{array}{c} \cos \omega t \\ \sin \omega t \end{array} \right)$ The forces $\vec F_\text{cf}$ and $- \vec F_\text{cf}$ at the points $A$ and $B$ exert a torque $\vec T$ on the mobile phone: $\vec T = b \vec e_y \times \vec F_\text{cf} + (-b \vec e_y) \times (-\vec F_\text{cf}) = -\frac{8 m \omega^2 a b}{3 \pi } \cos (\omega t) \cdot \vec e_z$ The rotational movement of the mobile phone obeys the equation $I \ddot \theta \vec e_z = \vec T$ with the moment of inertia $I$ , which is calculated by integration over the rectangular area of the mobile phone $\begin{aligned} I &= \int r^2 dM = \frac{M}{w l} \int_{-l/2}^{l/2} \int_{-w/2}^{w/2} (x^2 + y^2) dx dy \\ &= \frac{M}{w l} \left( \int_{-w/2}^{w/2} x^2 dx \cdot \int_{-l/2}^{l/2} dy + \int_{-w/2}^{w/2} dx \cdot \int_{-l/2}^{l/2} y^2 dy \right)\\ &= \frac{M}{w l} \left( \frac{2}{3} \left( \frac{w}{2} \right)^3 \cdot l + w \cdot \frac{2}{3} \left( \frac{l}{2} \right)^3 \right) =\frac{M }{12} (w^2 + l^2) \end{aligned}$ The equation of motion for the rotation angle yields: $\begin{aligned} & & \frac{d^2}{dt^2} \theta(t) &= \frac{T(t)}{I} = - \frac{32}{\pi} \frac{m}{M} \frac{ab}{w^2 + l^2} \omega^2 \cos (\omega t) \\ \Rightarrow & & \theta(t) &= \frac{32}{\pi} \frac{m}{M} \frac{ab}{w^2 + l^2} \cos (\omega t) \\ \Rightarrow & & A &= \frac{32}{\pi} \frac{m}{M} \frac{ab}{w^2 + l^2} \approx 6 \cdot 10^{-5} \end{aligned}$